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a(n) = floor(9*n/8).
6

%I #22 Sep 08 2022 08:46:10

%S 0,1,2,3,4,5,6,7,9,10,11,12,13,14,15,16,18,19,20,21,22,23,24,25,27,28,

%T 29,30,31,32,33,34,36,37,38,39,40,41,42,43,45,46,47,48,49,50,51,52,54,

%U 55,56,57,58,59,60,61,63,64,65,66,67,68,69,70,72,73,74,75,76,77,78,79,81,82,83,84,85,86,87,88,90,91,92,93,94,95

%N a(n) = floor(9*n/8).

%C Also: numbers not congruent to 8 (mod 9), or numbers whose base-9 expansion does not end in the digit "8".

%C Paz proves that for all n>0 there is a prime in Breusch's interval [n; a(n+3)], cf A248371.

%H G. A. Paz, <a href="http://emis.impa.br/EMIS/journals/GMN/yahoo_site_admin/assets/docs/1_GMN-2642-V15N1.120120222.pdf">On the Interval [n; 2n]: Primes, Composites and Perfect Powers</a>, Gen. Math. Notes 15 no. 1 (2013), 1-15.

%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,0,0,0,1,-1).

%F G.f.: x*(1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + 2*x^7) / ((1 + x)*(1 - x)^2*(1 + x^2)*(1 + x^4)). [_Bruno Berselli_, Oct 06 2014]

%F a(n) = n + floor(n/8) = a(n-1) + a(n-8) - a(n-9). [_Bruno Berselli_, Oct 06 2014]

%F a(n) = A168183(n+1) - 1. - _Philippe Deléham_, Dec 05 2013

%t Table[Floor[9 n/8], {n, 0, 90}] (* _Bruno Berselli_, Oct 06 2014 *)

%o (PARI) a(n)=9*n\8

%o (Magma) [Floor(9*n/8): n in [0..90]]; // _Bruno Berselli_, Oct 06 2014

%Y Cf. A032766, A004773, A001068, A047226, A047368, A004777, A168183, A281899 (subsequence).

%K nonn,easy

%O 0,3

%A _M. F. Hasler_, Oct 05 2014